矩阵求逆的c语言程序
//源程序如下# includesdio.h
#includeconio.h
#includestring.h
#includeiostream.h
#includestdlib.h
#包含math.h
# define max 100 void input style(int *)；//输入功能
void input(int **，int)；//输入功能
long danx(int **，int)；
int sgnx(int)；
void martx(int **，int)；int main(void)
{
int style=0，I = 0；
Int矩阵用C语言编程求解逆矩阵。 # include stdio . h # include stdlib . h # include malloc . h oid matrix opp(double * A，int m，int n，double * inv mat)；void MatrixInver(double *A，int m，int n，double * invmat)；双盈(双A[]，int m，int n)；int matrix_inv(double* p，int num，double * inv mat)；void MatrixOpp(double A[]，int m，int n，double * inv mat){ int I，j，x，y，k；double *SP = NULL，*AB = NULL，*B = NULL，X；SP =(double *)malloc(m * n * sizeof(double))；AB =(double *)malloc(m * n * sizeof(double))；b =(double *)malloc(m * n * sizeof(double))；X =剩余(A，m，n)；X = 1/X；for(I = 0；我是m；I)for(j = 0；j n；j){ for(k = 0；k m * n；k)B[k]= A[k]；{ for(x = 0；x n；x)B[I * n x]= 0；for(y = 0；y m；y)B[m * y j]= 0；b[I * n j]= 1；SP[i * n j] =盈余(B，m，n)；AB[I * n j]= X * SP[I * n j]；} } matrix inver(AB，m，n，inv mat)；免费(SP)；免费(AB)；免费(B)；}void MatrixInver(double A[]，int m，int n，double * invmat){ int I，j；double * B = invmatfor(I = 0；I n；I)for(j = 0；j m；j)B[I * m j]= A[j * n I]；}double盈余(double A[]，int m，int n){ int I，j，k，p，r；double X，temp = 1，temp1 = 1，s = 0，S1 = 0；if(n = = 2){ for(I = 0；我是m；I)for(j = 0；j n；j)if((I j)% 2)temp 1 * = A[I * n j]；else temp * = A[I * n j]；X =温度-温度1；} else { for(k = 0；k n；k){ for(I = 0，j = k；i m，j n；I，j)temp * = A[I * n j]；if(m-I){ for(p = m-I，r = m-1；p 0；p -，r-)temp * = A[r * n p-1]；} s = temptemp = 1；} for(k = n-1；k = 0；k-){ for(I = 0，j = k；i m，j = 0；I，j-)temp 1 * = A[I * n j]；if(m-I){ for(p = m-1，r = I；r m；p -，r)temp 1 * = A[r * n p]；} s1 = temp1temp1 = 1} X = s-S1；}返回X；}int matrix_inv(double* p，int num，double * invmat){ if(p = = NULL | | invmat = = NULL){ return 1；} if(num 10){ return 2；} matrix opp(p，num，num，inv mat)；返回0；} int main(){ int I，j；int numdouble * arr = NULLdouble * result = NULLint标志；Printf("请输入矩阵维度:\ n ")；scanf("%d "，数字)；arr =(double *)malloc(sizeof(double)* num * num)；result =(double *)malloc(sizeof(double)* num * num)；Printf("请输入%d*%d矩阵:\n "，num，num)；for(I = 0；i numI){ for(j = 0；j numj){ scanf(" % lf "，arr[I * num j])；} } flag = matrix _ inv(arr，num，result)；If (flag = = 0) {printf("逆矩阵为:\ n ")；for(I = 0；i num * numI){ printf(" % lf \ t " ,*(结果I))；if(I % num = =(num-1))printf(" \ n ")；}} else if (flag = = 1) {printf ("p/q为空\ n ")；} else { printf("超过最大尺寸\ n ")；}系统(“暂停”)；免费(arr)；免费(结果)；返回0；}
最后更新于 2023-10-11 07:52:43 并被添加「」标签，已有 位童鞋阅读过。
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